Đề tài " The kissing number in four dimensions " pdf

Annals of Mathematics The kissing number in four dimensions By Oleg R Musin Annals of Mathematics, 168 (2008), 1–32 The kissing number in four dimensions By Oleg R Musin Abstract The kissing number problem asks for the maximal number k(n) of equal size nonoverlapping spheres in n-dimensional space that can touch another sphere of the same size This problem in dimension three was the subject of a famous discussion between Isaac Newton and David Gregory in 1694 In three dimensions the problem was nally solved only in 1953 by Schătte and van der u Waerden In this paper we present a solution of a long-standing problem about the kissing number in four dimensions Namely, the equality k(4) = 24 is proved The proof is based on a modification of Delsarte’s method Introduction The kissing number k(n) is the highest number of equal nonoverlapping spheres in Rn that can touch another sphere of the same size In three dimensions the kissing number problem is asking how many white billiard balls can kiss (touch) a black ball The most symmetrical configuration, 12 billiard balls around another, is if the 12 balls are placed at positions corresponding to the vertices of a regular icosahedron concentric with the central ball However, these 12 outer balls not kiss each other and may all move freely So perhaps if you moved all of them to one side a 13th ball would possibly fit in? This problem was the subject of a famous discussion between Isaac Newton and David Gregory in 1694 It is commonly said that Newton believed the answer was 12 balls, while Gregory thought that 13 might be possible However, Casselman [8] found some puzzling features in this story The Newton-Gregory problem is often called the thirteen spheres problem Hoppe [18] thought he had solved the problem in 1874 However, there was a mistake — an analysis of this mistake was published by Hales [17] in 1994 Finally, this problem was solved by Schătte and van der Waerden in 1953 [31] u OLEG R MUSIN A subsequent two-page sketch of a proof was given by Leech [22] in 1956 The thirteen spheres problem continues to be of interest, and several new proofs have been published in the last few years [20], [24], [6], [1], [26] Note that k(4) ≥ 24 Indeed, the unit sphere in R4 centered at (0, 0, 0, 0) √ √ has 24 unit spheres around it, centered at the points (± 2, ± 2, 0, 0), with any choice of signs and any ordering of the coordinates The convex hull of these 24 points yields a famous 4-dimensional regular polytope - the “24-cell” Its facets are 24 regular octahedra Coxeter proposed upper bounds on k(n) in 1963 [10]; for n = 4, 5, 6, 7, and these bounds were 26, 48, 85, 146, and 244, respectively Coxeter’s bounds are based on the conjecture that equal size spherical caps on a sphere can be packed no denser than packing where the Delaunay triangulation with vertices at the centers of caps consists of regular simplices This conjecture was proved by Bărăczky in 1978 [5] oo The main progress in the kissing number problem in high dimensions was made at the end of the 1970s In 1978: Kabatiansky and Levenshtein found an asymptotic upper bound 20.401n(1+o(1)) for k(n) [21] (Currently known, the lower bound is 20.2075n(1+o(1)) [32].) In 1979: Levenshtein [23], and independently Odlyzko and Sloane [27] (= [9, Chap.13]), using Delsarte’s method, proved that k(8) = 240, and k(24) = 196560 This proof is surprisingly short, clean, and technically easier than all proofs in three dimensions However, n = 8, 24 are the only dimensions in which this method gives a precise result For other dimensions (for instance, n = 3, 4) the upper bounds exceed the lower In [27] the Delsarte method was applied in dimensions up to 24 (see [9, Table 1.5]) For comparison with the values of Coxeter’s bounds on k(n) for n = 4, 5, 6, 7, and this method gives 25, 46, 82, 140, and 240, respectively (For n = Coxeter’s and Delsarte’s methods only gave k(3) ≤ 13 [10], [27].) Improvements in the upper bounds on kissing numbers (for n < 24) were rather weak during the next years (see [9, Preface, Third Edition] for a brief review and references) Arestov and Babenko [2] proved that the bound k(4) ≤ 25 cannot be improved using Delsarte’s method Hsiang [19] claims a proof of k(4) = 24 His work has not yet received a positive peer review If M unit spheres kiss the unit sphere in Rn , then the set of kissing points is an arrangement on the central sphere such that the (Euclidean) distance between any two points is at least So the kissing number problem can be stated in another way: How many points can be placed on the surface of Sn−1 so that the angular separation between any two points is at least π/3? This leads to an important generalization: a finite subset X of Sn−1 is called a spherical ψ-code if for every pair (x, y) of X the inner product x · y ≤ cos ψ; i.e., the minimal angular separation is at least ψ Spherical codes have many applications The main application outside mathematics is in the design THE KISSING NUMBER IN FOUR DIMENSIONS of signals for data transmission and storage There are interesting applications to the numerical evaluation of n-dimensional integrals [9, Chap 3] Delsarte’s method (also known in coding theory as Delsarte’s linear programming method or Delsarte’s scheme) is widely used for finding bounds for codes This method is described in [9], [21] (see also [28] for a beautiful exposition) In this paper we present an extension of the Delsarte method that allowed us to prove the bound k(4) < 25, i.e k(4) = 24 This extension yields also a proof for k(3) < 13 [26] The first version of these proofs used numerical solutions of some nonconvex constrained optimization problems [25] (see also [28]) Now, using a geometric approach, we reduced it to relatively simple computations The paper is organized as follows: Section shows that the main theorem: k(4) = 24 easily follows from two lemmas: Lemma A and Lemma B Section reviews the Delsarte method and gives a proof of Lemma A Section extends Delsarte’s bounds and reduces the upper bound problem for ψ-codes to some optimization problem Section reduces the dimension of the corresponding optimization problem Section develops a numerical method for a solution of this optimization problem and gives a proof of Lemma B Acknowledgment I wish to thank Eiichi Bannai, Dmitry Leshchiner, Sergei Ovchinnikov, Makoto Tagami, Gănter Ziegler, and especially anonyu mous referees of this paper for helpful discussions and useful comments I am very grateful to Ivan Dynnikov who pointed out a gap in arguments in an earlier draft of [25] The main theorem Let us introduce the following polynomial of degree nine:1 f4 (t) := 1344 2688 1764 2048 1229 516 217 t − t + t + t − t − t − t− 25 25 25 125 125 125 500 125 Lemma A Let X = {x1 , , xM } be points in the unit sphere S3 Then M M f4 (xi · xj ) ≥ M S(X) = i=1 j=1 We give a proof of Lemma A in the next section The polynomial f4 was found by the linear programming method (see details in the appendix) This method for n = 4, z = 1/2, d = 9, N = 2000, t0 = 0.6058 gives E ≈ 24.7895 For f4 , coefficients were changed to “better looking” ones with E ≈ 24.8644 OLEG R MUSIN Lemma B Suppose X = {x1 , , xM } is a subset of S3 such that the angular separation between any two distinct points xi , xj is at least π/3 Then M M f4 (xi · xj ) < 25M S(X) = i=1 j=1 A proof of Lemma B is given at the end of Section k(4) = 24 Main theorem Proof Let X be a spherical π/3-code in S3 with M = k(4) points Then X satisfies the assumptions in Lemmas A, B Therefore, M ≤ S(X) < 25M From this M < 25 follows, i.e M ≤ 24 From the other side we have k(4) ≥ 24, showing that M = k(4) = 24 Delsarte’s method From here on we will speak of x ∈ Sn−1 , alternatively, of points in Sn−1 or of vectors in Rn Let X = {x1 , x2 , , xM } be any finite subset of the unit sphere Sn−1 ⊂ Rn , Sn−1 = {x : x ∈ Rn , x · x = ||x||2 = 1} By φi,j = dist(xi , xj ) we denote the spherical (angular) distance between xi , xj Clearly, cos φi,j = xi · xj 3-A Schoenberg’s theorem Then || Let u1 , u2 , , uM be any real numbers ui xi ||2 = cos φi,j ui uj ≥ 0, i,j or equivalently the Gram matrix cos φi,j is positive semidefinite (n) Schoenberg [29] extended this property to Gegenbauer polynomials Gk (n) He proved: The matrix Gk (cos φi,j ) is positive semidefinite for any finite X ⊂ Sn−1 Schoenberg proved also that the converse holds: If f (t) is a real polynomial and for any finite X ⊂ Sn−1 the matrix f (cos φi,j ) is positive semidefinite, (n) then f (t) is a linear combination of Gk (t) with nonnegative coefficients 3-B The Gegenbauer polynomials Let us recall definitions of Gegenbauer (n) polynomials Ck (t), which are defined by the expansion ∞ (1 − 2rt + r2 )(2−n)/2 = (n) rk Ck (t) k=0 THE KISSING NUMBER IN FOUR DIMENSIONS (n) (n) (n) Then the polynomials Gk (t) := Ck (t)/Ck (1) are called Gegenbauer or (n) ultraspherical polynomials (So the normalization of Gk is determined by the (n) (n) condition Gk (1) = 1.) Also the Gegenbauer polynomials Gk can be defined by the recurrence formula: (n) (n) G0 (n) = 1, G1 (n) = t, , Gk = (n) (2k + n − 4) t Gk−1 − (k − 1) Gk−2 k+n−3 They are orthogonal on the interval [−1, 1] with respect to the weight function ρ(t) = (1 − t2 )(n−3)/2 (see details in [7], [9], [15], [29]) In the case (n) (4) n = 3, Gk are Legendre polynomials Pk , and Gk are Chebyshev polynomials of the second kind (but with a different normalization than usual, Uk (1) = 1), (4) Gk (t) = Uk (t) = sin ((k + 1)φ) , (k + 1) sin φ t = cos φ, k = 0, 1, 2, For instance, U0 = 1, U1 = t, U2 = (4t2 − 1)/3, U3 = 2t3 − t, U4 = (16t4 − 12t2 + 1)/5, , U9 = (256t9 − 512t7 + 336t5 − 80t3 + 5t)/5 3-C Delsarte’s inequality If a symmetric matrix is positive semidefinite, then the sum of all its entries is nonnegative Schoenberg’s theorem implies (n) that the matrix Gk (ti,j ) is positive semidefinite, where ti,j := cos φi,j , Then M M (n) Gk (ti,j ) ≥ (3.1) i=1 j=1 Definition We denote by G+ the set of continuous functions f : [−1, 1] n → R representable as series ∞ (n) ck Gk (t) f (t) = k=0 whose coefficients satisfy the following conditions: ∞ ck ≥ for k = 1, 2, , c0 > 0, ck < ∞ f (1) = k=0 Suppose f ∈ G+ and let n M M S(X) = Sf (X) := f (ti,j ) i=1 j=1 Using (3.1), we get ∞ S(X) = k=0 ⎛ ck ⎝ M M i=1 j=1 ⎞ (n) Gk (ti,j )⎠ M M (n) ≥ c0 G0 (ti,j ) = c0 M i=1 j=1 OLEG R MUSIN Then S(X) ≥ c0 M (3.2) (4) 3-D Proof of Lemma A The expansion of f4 in terms of Uk = Gk is 153 871 128 21 U2 + U3 + U4 + U9 25 250 25 20 We see that f4 ∈ G+ with c0 = So Lemma A follows from (3.2) f4 = U0 + U1 + 3-E Delsarte’s bound Let X = {x1 , , xM } ⊂ Sn−1 be a spherical ψ-code, i.e for all i = j, ti,j = cos φi,j = xi · xj ≤ z := cos ψ, i.e ti,j ∈ [−1, z] (but ti,i = 1) Suppose f ∈ G+ and f (t) ≤ for all t ∈ [−1, z]; then f (ti,j ) ≤ for all n i = j That implies Sf (X) = M f (1) + 2f (t1,2 ) + + 2f (tM −1,M ) ≤ M f (1) If we combine this with (3.2), then we get M ≤ f (1)/c0 Let A(n, ψ) be the maximal size of a ψ-code in Sn−1 Then we have: f (1) c0 The inequality (3.3) plays a crucial role in the Delsarte method (see details in [2], [3],[4], [9], [13], [14], [21], [23], [27]) If z = 1/2 and c0 = 1, then (3.3) implies k(n) = A(n, π/3) ≤ f (1) (3.3) A(n, ψ) ≤ Levenshtein [23], and independently Odlyzko and Sloane [27] for n = 8, 24 have found suitable polynomials f (t): f (t) ≤ for all t ∈ [−1, 1/2], f ∈ G+ , c0 = n with f (1) = 240 for n = 8; and f (1) = 196560 for n = 24 Then k(8) ≤ 240, k(24) ≤ 196560 For n = 8, 24 the minimal vectors in sphere packings E8 and Leech lattice give these kissing numbers Thus k(8) = 240, and k(24) = 196560 When n = 4, a polynomial f of degree with f (1) ≈ 25.5585 was found in [27] This implies 24 ≤ k(4) ≤ 25 An extension of Delsarte’s method 4-A An extension of Delsarte’s bound Let f (t) be any real function on the interval [−1, 1] Let, for a given ψ, z := cos ψ Consider on the sphere Sn−1 points y0 , y1 , , ym such that (4.1) yi · yj ≤ z for all i = j, f (y0 · yi ) > for ≤ i ≤ m THE KISSING NUMBER IN FOUR DIMENSIONS Definition For fixed y0 ∈ Sn−1 , m ≥ 0, z, and f (t) let us define the family Qm (y0 ) = Qm (y0 , n, f ) of finite sets of points from Sn−1 by the formula {y0 }, m = 0, {Y = {y1 , , ym } ⊂ Sn−1 : {y0 } ∪ Y satisfies (4.1)}, m ≥ Qm (y0 ) := Denote μ = μ(n, z, f ) := max{m : Qm (y0 ) = ∅} For ≤ m ≤ μ we define the function H = Hf on the family Qm (y0 ): H(y0 ) := f (1) for m = 0, H(y0 ; Y ) = H(y0 ; y1 , , ym ) := f (1) + f (y0 · y1 ) + + f (y0 · ym ) for m ≥ Let hm = hm (n, z, f ) := sup Y ∈Qm (y0 ) {H(y0 ; Y )}, hmax := max {h0 , h1 , , hμ } Theorem Suppose f ∈ G+ Then n A(n, ψ) ≤ hmax (n, cos ψ, f ) = max{h0 , h1 , , hμ } c0 c0 Proof Let X = {x1 , , xM } ⊂ Sn−1 be a spherical ψ-code Since f ∈ G+ , (3.2) yields: S(X) ≥ c0 M n Denote J(i) := {j : f (xi · xj ) > 0, j = i}, X(i) := {xj : j ∈ J(i)} Then M f (xi · xj ) ≤ f (1) + Si (X) := j=1 f (xi · xj ) = H(xi ; X(i)) ≤ hmax , j∈J(i) so that M (4.2) Si (X) ≤ M hmax S(X) = i=1 We have c0 M ≤ S(X) ≤ M hmax , i.e c0 M ≤ hmax as required Note that h0 = f (1) If f (t) ≤ for all t ∈ [−1, z], then μ(n, z, f ) = 0, i.e hmax = h0 = f (1) Therefore, this theorem yields the Delsarte bound M ≤ f (1)/c0 4-B The class of functions Φ(t0 , z) The problem of evaluating hmax in the general case looks even more complicated than the upper bound problem for spherical ψ-codes It is not clear how to find μ, which is an optimal arrangement for Y ? Here we consider this problem only for a very restrictive class of functions Φ(t0 , z) For the bound given by Theorem we need f ∈ G+ n However, for evaluations of hm we not need this assumption So we not assume that f ∈ G+ n OLEG R MUSIN Definition Let real numbers t0 , z satisfy > t0 > z ≥ We denote by Φ(t0 , z) the set of functions f : [−1, 1] → R such that f (t) ≤ for t ∈ [−t0 , z] Let f ∈ Φ(t0 , z), and let Y ∈ Qm (y0 , n, f ) Denote e0 := −y0 , θ0 := arccos t0 , θi := dist(e0 , yi ) for i = 1, , m (In other words, e0 is the antipodal point to y0 ) It is easy to see that f (y0 · yi ) > only if θi < θ0 Therefore, Y is a spherical ψ-code in the open spherical cap Cap(e0 , θ0 ) of center e0 and radius θ0 with π/2 ≥ ψ > θ0 This assumption is quite restrictive and in particular derives the convexity property for Y We use this property in the next section 4-C Convexity property A subset of Sn−1 is called spherically convex if it contains, with every two nonantipodal points, the small arc of the great circle containing them The closure of a convex set is convex and is the intersection of closed hemispheres (see details in [12]) Let Y = {y1 , , ym } ⊂ Cap(e0 , θ0 ), θ0 < π/2 Then the convex hull of Y is well defined, and is the intersection of all convex sets containing Y Denote the convex hull of Y by Δm = Δm (Y ) Recall a definition of a vertex of a convex set: A point y ∈ W is called the vertex (extremal point) of a spherically convex closed set W , if the set W \ {y} is spherically convex or, equivalently, there are no points x, z from W for which y is an interior point of the minor arc xz of large radius connecting x, z Theorem Let Y = {y1 , , ym } ⊂ Sn−1 be a spherical ψ-code Suppose Y ⊂ Cap(e0 , θ0 ), and < θ0 < ψ ≤ π/2 Then any yk is a vertex of Δm Proof The cases m = 1, are evident For the case m = the theorem can be easily proved by contradiction Indeed, suppose that some point, for instance, y2 , is not a vertex of Δ3 Then, firstly, the set Δ3 is the arc y1 y3 , and, secondly, the point y2 lies on the arc y1 y3 From this it follows that dist(y1 , y3 ) ≥ 2ψ, since Y is a ψ-code On the other hand, according to the triangle inequality, we have 2ψ ≤ dist(y1 , y3 ) ≤ dist(e0 , y1 ) + dist(e0 , y3 ) < 2θ0 We obtained the contradiction It remains to prove the theorem for m ≥ In this paper we need only one fact from spherical trigonometry, namely the law of cosines (or the cosine theorem): cos φ = cos θ1 cos θ2 + sin θ1 sin θ2 cos ϕ, where for a spherical triangle ABC the angular lengths of its sides are dist(A, B) = θ1 , dist(A, C) = θ2 , dist(B, C) = φ, and ∠BAC = ϕ THE KISSING NUMBER IN FOUR DIMENSIONS By the assumptions: θk = dist(yk , e0 ) < θ0 < ψ for ≤ k ≤ m; φk,j := dist(yk , yj ) ≥ ψ, k = j Let us prove that there is no point yk belonging both to the interior of Δm and relative interior of some facet of dimension d, ≤ d ≤ dim Δm Assume the converse Then consider the great (n − 2)-sphere Ωk such that yk ∈ Ωk , and Ωk is orthogonal to the arc e0 yk (Note that θk > Conversely, yk = e0 and φk,j = θj ≤ θ0 < ψ.) The great sphere Ωk divides Sn−1 into two closed hemispheres: H1 and H2 Suppose e0 lies in the interior of H1 , then at least one yj belongs to H2 Consider the triangle e0 yk yj and denote by γk,j the angle ∠e0 yk yj in this triangle The law of cosines yields cos θj = cos θk cos φkj + sin θk sin φk,j cos γk,j Since yj ∈ H2 , we have γk,j ≥ 90◦ , and cos γk,j ≤ (Fig 1) From the conditions of Theorem there follow the inequalities sin θk > 0, sin φk,j > 0, cos θk > 0, cos θj > yj r     H2   yk     r r H1 Ωk e0 Figure Hence, using the cosine theorem we obtain cos θj = cos θk cos φk,j + sin θk sin φk,j cos γk,j , < cos θj ≤ cos θk cos φk,j From these inequalities and < cos θk < it follows that, firstly, < cos φk,j i.e ψ ≤ φk,j < π/2 , and, secondly, the inequalities cos θj < cos φk,j ≤ cos ψ Therefore, θj > ψ This contradiction completes the proof of Theorem 18 OLEG R MUSIN for k = 4, 5, and does not change θi It can be shown in an elementary way that there is a turn such that φ2,4 or φ2,5 becomes equal to ψ, a contradiction In three dimensions there exist only two combinatorial types of convex polytopes with five vertices: (A) and (B) (see Fig 5) In the case (A) the arc y3 y5 lies inside Δ5 , and for (B): y2 y3 y4 y5 is a facet of Δ5 y y s € h €€€ €€sy3  h pp p  h  p p ¡ y4 s h p p p ¡  p p p p ¡  f p ph pp p  p p p f h ¡ pp p fh  p p ¡ p p pˆ fh ¡   y sp p ˆ ˆˆ ˆˆˆ fh ¡ f¡ ˆhs y2 s d ¡h ¡ h d ¡ h d d ¡ y4 s h € ¡   h€€€d €€sy3 h d ¡   h ¡    h   s   y5 ¡ €€ h   €€ €€hs y   (A) (B) Figure (2) By sij we denote the arc yi yj , and by sijk denote the triangle yi yj yk Let sijk be the intersection of the great 2−hemisphere Qijk and Δ5 , where ˜ Qijk contains yi , yj , yk and is bounded by the great circle passing through ˜ yi , yj Proposition yields: if there are no i, j, k such that e0 ∈ sijk , then deg yi ≥ for all i It remains to consider all cases e0 ∈ sijk Note that for (A), sijk = sijk only ˜ ˜ for three cases, i = 1, 2, 4; where j = 3, k = 5, or j = 5, k = (˜i35 = si53 ) s ˜ (3) Lemma yields that deg yk > Now we consider the cases deg yk = 1, If deg yk = 1, φk, = ψ, then e0 ∈ sk Indeed, otherwise there exists the great circle Ω in S3 such that Ω contains y , and the great sphere passes through Ω and yk does not pass through e0 Then Lemma implies that a rotation R(ϕ, Ω) of yk with sufficiently small ϕ decreases θk — a contradiction Since θ0 < ψ, e0 cannot be a vertex of Δ5 Therefore, e0 lies inside sk From this we have: If sij for any j does not intersect sk , then deg yi ≥ Arguing as above, we can prove that If deg yk = 2, φk,i = φk,j = ψ, then e0 ∈ sijk ˜ (4) Now we prove that deg yk ≥ for all k Conversely, deg yk = 1, e0 ∈ sk a) First we consider the case when sk is an “external” edge of Δ5 For type (A) this means sk differs from s35 , and for (B) it is not s35 or s24 Since Δ5 is convex, there exists the great 2-sphere Ω2 passes through yk , y such that three other points yi , yj , yq lie inside the hemisphere H+ bounded by Ω2 Let Ω THE KISSING NUMBER IN FOUR DIMENSIONS 19 be the great circle in Ω2 that contains y and is orthogonal to the arc sk Then (Lemma 3) there exists a small turn of yi , yj , yq around Ω that simultaneously decreases θi , θj , θq — a contradiction b) For type (A) when deg y3 = 1, φ3,5 = ψ, e0 ∈ s35 ; we claim that s124 is a regular triangle with side length ψ Indeed, from a) it follows that deg yi ≥ for i = 1, 2, Moreover, if deg yi = 2, then e0 = s35 s124 Therefore, in any case, φ1,2 = φ1,4 = φ2,4 = ψ We have the arc s35 and the regular triangle s124 , both are with edge lengths ψ Then from Lemma it follows that some φi,j < ψ — a contradiction c) Now for type (B) consider the case: deg y3 = 1, φ3,5 = ψ, e0 ∈ s35 Then for y2 we have: deg y2 = only if φ2,4 = ψ, e0 = s24 s35 ; deg y2 = only if φ2,4 = φ2,5 = ψ; and φ2,4 = φ1,2 = φ2,5 = ψ if deg y2 = Thus, in any case, φ2,4 = ψ We have two intersecting diagonals s24 , s35 of lengths ψ Then Lemma contradicts the assumption that Y is a ψ-code This contradiction concludes the proof that deg yk ≥ for all k (5) Finally we prove that deg yk ≥ for all k Assume the converse Then deg yk = 2, e0 ∈ sijk , where φk,i = φk,j = ψ ˜ Case facet Let sijk be a facet of Δ5 , and e0 ∈ sij By the same argument / as in (4a), where Ω2 the great sphere contains sijk , and Ω the great circle passes through yi , yj , we can prove that there exist shift decreases θ , θq for two other points y , yq from Y , a contradiction If e0 ∈ sij , then any turn of s q around Ω does not change θ and θq However, if this turn is in a positive direction, then it decreases φk, and φk,q Clearly, there exists a turn when φk, or φk,q is equal to ψ — a contradiction It remains to consider all cases where sijk is not a facet These are: s124 , s135 (type (A) and type (B)), s234 (type (B)) Case s124 We have deg y1 = 2, φ1,2 = φ1,4 = ψ, e0 ∈ s124 Consider a small turn of y3 around s24 towards y1 If e0 ∈ s24 , then this turn decreases / θ3 Therefore, the irreducibility yields φ3,5 = ψ In the case e0 ∈ s24 , θ3 = θ3 , but φ1,3 decreases This again implies φ3,5 = ψ Since s35 cannot intersects a regular triangle s124 [see Lemma 2, (4b)], φ2,4 > ψ Then deg y2 = deg y4 = (Since e0 ∈ s124 , deg y2 = only if φ2,4 = ψ.) Thus we have three isosceles triangles s243 , s241 , s245 Using this and φ3,5 = ψ, we obviously have φ1,i < ψ for i = 3, 5, — a contradiction Case s135 (type (B)) is equivalent to the Case s124 ˜ ˜ Case s135 (type (A)) This case has two subcases: s351 , s153 In the subcase s135 we have deg y1 = 2, φ1,3 = φ1,5 = ψ, e0 ∈ s135 If e0 ∈ s135 , then any turn ˜ ˜ / of y1 around s35 decreases θ1 (Lemma 3) Then e0 ∈ s135 Clearly, any small turn of y2 around s35 increases φ2,4 On the other hand, this turn decreases θ2 20 OLEG R MUSIN (if e0 ∈ s35 ) and φ1,2 Arguing as above, we get a contradiction The subcase / s315 , where φ3,5 = ψ, can be proven by the same arguments as Case s124 ˜ Case s234 (type (B)) This case has two subcases: s243 , s234 It is not hard ˜ ˜ ˜ to see that s243 follows from Case facet, and s234 can be proven in the same ˜ way as subcase s135 This concludes the proof ˜ Lemma yields that the degree of any vertex of Γψ (Y ) is not less than This implies that at least one vertex of Γψ (Y ) has degree Indeed, if all vertices of Γψ (Y ) are of degree 3, then the sum of the degrees equals 15, i.e is not an even number There exists only one type of Γψ (Y ) with these conditions (Fig 6) The lengths of all edges of Δ5 except y2 y4 , y3 y5 are equal to ψ For fixed φ2,4 = α, Δ5 is uniquely defined up to isometry Therefore, we have the 1-parametric family P5 (α) on S3 If φ3,5 ≥ φ2,4 , then z ≥ cos α ≥ 2z − y t1 € h €€€ €€ty3  h  h  ¡   y4 t h   ¡ pp   ppp h ¡ pp h   ¡ pp pph   ¡  α pp hp ¡   y5 t ˆ p ˆ ˆ php ¡ ˆˆˆ ˆˆphpt y p ¡ Figure 6: P5 (α) Thus Theorem 4, Corollary and Lemma for n = yield: Theorem Let Y ⊂ S3 be an irreducible set, |Y | = m ≤ Then Δm for ≤ m ≤ is a regular simplex of edge lengths ψ, and Δ5 is isometric to P5 (α) for some α ∈ [ψ, arccos (2z − 1)] 5-F Optimization problem We see that if Y is optimal, then for some cases Y can be determined up to isometry For fixed yi ∈ Sn−1 , i = 1, , m, the function H depends only on a position y = −y0 = e0 ∈ Sn−1 Now, Hm (y) := f (1) + f (−y · y1 ) + + f (−y · ym ); i.e Hm (y) = H(−y; Y ) Thus for hm we have the following (n − 1)-dimensional optimization problem: hm = max {Hm (y)} y subject to the constraint y ∈ T (Y, θ0 ) := {y ∈ Δm ⊂ Sn−1 : y · yi ≥ t0 , i = 1, , m} THE KISSING NUMBER IN FOUR DIMENSIONS 21 We present an efficient numerical method for solving this problem in the next section On calculations of hm In this technical section we explain how to find an upper bound on hm for n = 4, m ≤ Note that Theorem gives for computation of hm a low-dimensional optimization problem (see 5-F) Our first approach for this problem was to apply numerical methods [25] However, that is a nonconvex constrained optimization problem In this case, the Nelder-Mead simplex method and other local improvements cannot guarantee finding a global optimum It is possible (using estimations of derivatives) to organize the computational process in such way that it gives a global optimum However, such solutions are very hard to verify and some mathematicians not accept that kind of proof Fortunately, using a geometric approach, estimations of hm can be reduced to relatively simple computations ˜ Throughout this section we use the function f (θ) defined for f ∈ Φ∗ (z) by ˜ f (θ) := f (− cos θ) −∞ ≤ θ ≤ θ0 = arccos t0 (see Definition 4) θ > θ0 ˜ Since f ∈ Φ∗ (z), f (θ) is a monotone decreasing function in θ on [0, θ0 ] 6-A The case m = Suppose m = and Y is optimal for f ∈ Φ∗ (z) Then Δ2 = y1 y2 is an arc of length ψ, e0 ∈ Δ2 , and θ1 + θ2 = ψ, where θi ≤ θ0 ˜ ˜ (see Lemma and (5.2)) The efficient function F (θ1 , θ2 ) = f (1)+ f (θ1 )+ f (θ2 ) is a symmetric function in θ1 , θ2 We can assume that θ1 ≤ θ2 , and then θ1 ∈ [ψ − θ0 , ψ/2] Since Θ2 (θ1 ) := ˜ ψ − θ1 is a monotone decreasing function, f (Θ2 (θ1 )) is a monotone increasing function in θ1 Thus for any θ1 ∈ [u, v] ⊂ [ψ − θ0 , ψ/2] we have ˜ ˜ F (θ1 , θ2 ) ≤ Φ2 ([u, v]) := f (1) + f (u) + f (ψ − v) Let u1 = ψ − θ0 , u2 , , uN , uN +1 = ψ/2 be points in [ψ − θ0 , ψ/2] such that ui+1 = ui + ε, where ε = (θ0 − ψ/2)/N If θ1 ∈ [ui , ui+1 ], then h2 = H(y0 ; Y ) = F (θ1 , θ2 ) ≤ Φ2 ([ui , ui+1 ]) Thus h2 ≤ λ2 (N, ψ, θ0 ) := max {Φ2 (si )}, where si := [ui , ui+1 ] 1≤i≤N Clearly, λ2 (N, ψ, θ0 ) tends to h2 as N → ∞ (ε → 0) This implies a very simple method for calculation of h2 Now we extend this approach to higher m 6-B The function Θk Suppose we know that (up to isometry) optimal Y = {y1 , , ym } ⊂ Sn−1 Let us assume that dim Δm = n − 1, and V := convex hull of {y1 yn−1 } is a facet of Δm Then rank{y1 , , yn−1 } = n−1, 22 OLEG R MUSIN and Y belongs to the hemisphere H+ , where H+ contains Y and is bounded ˜ by the great sphere S passing through V Let us show that any y = y+ ∈ H+ is uniquely determined by the set of distances θi = dist(y, yi ), i = 1, , n − Indeed, there are at most two solutions: y+ ∈ H+ and y− ∈ H− of the quadratic equation (6.1) y · y = with y · yi = cos θi , i = 1, , n − ˜ Note that y+ = y− if and only if y ∈ S This implies that θk , k ≥ n, is determined by θi , i = 1, , n − 1; θk = Θk (θ1 , , θn−1 ) It is not hard to solve (6.1) and, therefore, to give an explicit expression for Θk For instance, let Δn be a regular simplex of edge lengths π/3 (We need this case for n = 3, 4.) Then3 cos θ3 = cos Θ3 (θ1 , θ2 ) = cos θ1 + cos θ2 + cos θ4 = cos Θ4 (θ1 , θ2 , θ3 ) = √ + 10 − 8[cos θ1 cos θ2 + (cos θ2 − cos θ1 )2 ] ; cos θ1 + cos θ2 + cos θ3 + cos θ1 cos θ2 + cos θ1 cos θ3 + cos θ2 cos θ3 − (cos2 θ1 + cos2 θ2 + cos2 θ3 ) 6-C Extremal points of Θk on D Let a = (a1 , , an−1 ), where < ≤ θ0 < ψ (Recall that φi,j = dist(yi , yj ); cos ψ = z; cos θ0 = t0 ) Now we consider a domain D(a) in H+ , where D(a) = {y ∈ H+ : dist(y, yi ) ≤ , ≤ i ≤ n − 1} In other words, D(a) is the intersection of the closed caps Cap(yi , ) in H+ : n−1 Cap(yi , ) D(a) = H+ i=1 Suppose dim D(a) = n − Then D(a) has “vertices”, “edges”, and “k-faces” for k ≤ n − Indeed, let σ ⊂ I := {1, , n − 1}, < |σ| ≤ n − 1; ˜ Fσ := {y ∈ D(a) : dist(y, yi ) = ∀ i ∈ σ} ˜ ˜ It is easy to prove that dim Fσ = n − − |σ|; Fσ belongs to the boundary B ˜σ ⊂ Fσ ˜ of D(a); and if σ ⊂ σ , then F I am very grateful to referees for these explicit formulas 23 THE KISSING NUMBER IN FOUR DIMENSIONS Now we consider the minimum of Θk (θ1 , , θn−1 ) on D(a) for k ≥ n In other words, we are looking for a point pk (a) ∈ D(a) such that dist(yk , pk (a)) = dist(yk , D(a)) Since φi,k ≥ ψ > θ0 , all yk lie outside D(a) Clearly, Θk achieves its minimum at some point in B Therefore, there is σ ⊂ I such that ˜ pk (a) ∈ Fσ (6.2) ˜ Suppose σ = I, then Fσ is a vertex of D(a) Let us denote this point by p∗ (a) Note that the function Θk at the point p∗ (a) is equal to Θk (a) Let σk (a) denote σ ⊂ I of the maximal size such that σ satisfies (6.2) Then for σk (a) = I, pk (a) = p∗ (a), and for |σk (a)| < n − 1, pk (a) belongs to ˜ the open part of Fσk (a) Consider n = There are two cases for pk (a) (see Fig 7): p3 (a) = ˜ p∗ (a) = F{1,2} , and p4 (a) is the intersection in H+ of the great circle passing ˜ ˜ through y1 , y4 , and the circle S(y1 , a1 ) of center y1 and radius a1 (F{1} ⊂ ˜ S(y1 , a1 )) The same holds for all dimensions Denote by Sσ (k) the great |σ|−dimensional sphere passing through yi , ˜ i ∈ σ, and yk Let S(yi , ) be the sphere of center yi and radius ; and for σ⊂I ˜ ˜ S(yi , ) Sσ := i∈σ ˜ Denote by s(σ, k) the intersection of Sσ (k) and Sσ in D(a), s(σ, k) = Sσ (k) p s y4 pp pp pp p∗ r (a) p p H+ pp ˜ F{2} p p pr p4 (a) pp ˜ p p D(a) F{1} p sp s ˜ Sσ D(a) y3 s y1 y2 Figure θ2 = a2 θ = b2 E(b, a) θ1 = a1 θ = b1 Figure Lemma Suppose D(a) = ∅, < ≤ θ0 for all i, and k ≥ n Then (i) pk (a) ∈ s(σk (a), k), (ii) if s(σ, k) = ∅, |σ| < n − 1, then s(σ, k) consists of the one point pk (a) 24 OLEG R MUSIN ∗ Proof (i) Let θk := Θk (pk (a)) = dist(yk , pk (a)) Since Θk achieves its ∗ ˜ ˜ minimum at pk (a), the sphere S(yk , θk ) touches the sphere Sσ(a) at pk (a) If some sphere touches the intersections of spheres, then the touching point belongs to the great sphere passing through the centers of these spheres Thus pk (a) ∈ Sσ(a) (k) (ii) Note that s(σ, k) belongs to the intersection in D(a) ⊂ H+ of the spheres S(yi , ), i ∈ σ, and Sσ (k) Any intersection of spheres is also a sphere Since ˜ dim Sσ (k) + dim Sσ = n − 1, this intersection is empty, or is a 0−dimensional sphere (i.e 2-points set) In the last case, one point lies in H+ , and another one in H− Therefore, s(σ, k) = ∅, or s(σ, k) = {p} Denote by σ the maximal size σ ⊃ σ such that ˜ ˜ s(σ , k) = {p} It is not hard to see that S(yk , dist(yk , p)) touches Sσ at p Thus p = pk (a) Lemma implies a simple method for calculations of the minimum of Θk on D(a) For this we can consider s(σ, k), σ ⊂ I, and if s(σ, k) = ∅, then s(σ, k) = {pk (a)}, so then Θk attains its minimum at this point In the case when Δn is a simplex we can find the minimum by a very simple method Corollary Suppose |Y | = n, < ≤ θ0 for all i, and D(a) lies inside Δn Then θn ≥ Θn (a1 , , an−1 ) for all y ∈ D(a) Proof Clearly, Δn is a simplex Since D(a) lies inside Δn , for |σ| < n − ˜ the intersection of Sσ and Sσ (k) is empty Thus pn (a) = p∗ (a) 6-D Upper bounds on Hm Suppose dim Δm = n − 1, and y1 yn−1 is a facet of Δm Then (see 5-F for the definitions of Hm and T (Y, θ0 )) ˜ Hm (y) = F (θ1 , , θn−1 , Θn , , Θm ) = Fm (θ1 , , θn−1 ), where ˜ ˜ ˜ ˜ Fm (θ1 , , θn−1 ) := f (1) + f (θ1 ) + + f (θn−1 ) + f (Θn (θ1 , , θn−1 )) ˜ + + f (Θm (θ1 , , θn−1 )) Lemma Suppose f ∈ Φ∗ (z), |Y | = m, dim Δm = n − 1, y1 yn−1 is a facet of Δm , dist(yi , yj ) ≥ ψ > θ0 for i = j, ≤ bi < ≤ θ0 for i = 1, , n − 1; and Θk (a) ≤ θ0 for all k ≥ n If D(a) = ∅, then Hm (y) ≤ ΦY (b, a) for any y ∈ E(b, a) := D(a) \ U (b), THE KISSING NUMBER IN FOUR DIMENSIONS 25 where ˜ ˜ ˜ ˜ ΦY (b, a) := f (1) + f (b1 ) + + f (bn−1 ) + f (Θn (pn (a))) + + f (Θm (pm (a))), n−1 U (b) := Cap(yi , bi ) i=1 Proof We have for ≤ i ≤ n − and y ∈ E(b, a), θi ≥ bi (Fig 8) By ˜ ˜ the monotonicity assumption this implies f (θi ) ≤ f (bi ) On the other hand, ˜ ˜ y ∈ D(a) Then Lemma yields f (θk ) ≤ f (Θk (pk (a))) for k ≥ n From Corollary and Lemma we obtain Corollary Let |Y | = n Suppose f, a, b, and Y satisfy the assumptions of Lemma and Corollary Then for any y ∈ E(b, a): ˜ ˜ ˜ Hm (y) ≤ f (1) + f (b1 ) + + f (bn−1 ) + f (Θn (a)) Let K(n, θ0 ) := [0, θ0 ]n−1 , i.e K(n, θ0 ) is an (n − 1)−dimensional cube of side length θ0 Consider for K(n, θ0 ) the cubic grid L(N ) of sidelength ε, where ε = θ0 /N for a given positive integer N Then the grid (tessellation) L(N ) consists of N n−1 cells, any cell c ∈ L(N ) (θ1 , , θn−1 ) in c we have bi (c) ≤ θi ≤ (c), (c) = bi (c) + ε, i = 1, , n − ˜ Let L(N ) be the subset of cells c in L(N ) such that D(a(c)) = ∅ There exists c ∈ L(N ) such that Hm attains its maximum on T (Y, θ0 ) at some point in E(b(c), a(c)) Therefore, Lemma yields Lemma Suppose f and Y satisfy the assumptions of Lemma 6, N is a positive integer, and y ∈ Δm is such that dist(y, yi ) ≤ θ0 for all i Then Hm (y) ≤ max {ΦY (b(c), a(c))} ˜ c∈L(N ) 6-E Upper bounds on hm Suppose Δm is a regular simplex of edge length ψ Then the efficient function F is a symmetric function in the variables θ1 , , θm Consider this problem only on the domain Λ := {y ∈ Δm : ψ − θ0 ≤ θ1 ≤ θ2 ≤ ≤ θm ≤ θ0 } ˜ Let LΛ (N ) be the subset of cells c in L(N ) such that E(b(c), a(c)) Λ = ∅ Then we have an explicit expression for Φm (c) := ΦY (b(c), a(c)) (see Corollary 6) For n = 4, Theorem implies that Δm is a regular simplex, where m = 2, 3, Thus from Lemma 7, hm ≤ λm (N, ψ, θ0 ) := max {Φm (c)} c∈LΛ (N ) Now we consider the case n = 4, m = Theorem yields: Δ5 is isometric to P5 (α) for some α ∈ [ψ, ψ := arccos (2z − 1)] (see Fig 6) Let the 26 OLEG R MUSIN vertices y1 , y2 , y3 of P5 (α) be fixed Then the vertices y4 (α), y5 (α) are uniquely determined by α Note that for any y ∈ D(θ0 , θ0 , θ0 ) the distance θ4 (α) := dist(y, y4 (α)) increases, and θ5 (α) decreases whenever α increases Let α1 = ψ, α2 , , αN , αN +1 = ψ be points in [ψ, ψ ] such that αi+1 = αi + , where = (ψ − ψ)/N Then θ4 (αi ) < θ4 (αi+1 ), θ5 (αi ) > θ5 (αi+1 ), so that ˜ ˜ f (θ4 (αi )) > f (θ4 (αi+1 )), ˜ ˜ f (θ5 (αi )) < f (θ5 (αi+1 )) Combining this with Lemma 7, we get h5 ≤ λ5 (N, ψ, θ0 ) := f (1) + max {R1,2,3 (c) + max {R4,5 (c, i)}}, ˜ c∈L(N ) 1≤i≤N ˜ ˜ ˜ R1,2,3 (c) = f (b1 (c)) + f (b2 (c)) + f (b3 (c)), ˜ ˜ R4,5 (c, i) = f (Θ4 (p4 (a(c), αi ))) + f (Θ5 (p5 (a(c), αi+1 ))), where pk (a, α) = pk (a) with yk = yk (α) Clearly, λm (2N, ψ, θ0 ) ≤ λm (N, ψ, θ0 ) It is not hard to show that hm ≤ λm (ψ, θ0 ) := lim λm (N, ψ, θ0 ) N →∞ Finally let us consider the case: n = 4, m = In this case, we give an upper bound on h6 by a separate argument √ √ Lemma Let n = 4, f ∈ Φ∗ (z), z > t0 > z, θ0 ∈ [arccos z, θ0 ] Then √ ˜ h6 ≤ max { f (θ0 ) + λ5 (ψ, θ0 ), f (− z) + λ5 (ψ, θ0 ) } Proof Let Y = {y1 , , y6 } ⊂ C(e0 , θ0 ) ⊂ S3 , where Y is an optimal z-code We may assume that θ1 ≤ θ2 ≤ ≤ θ6 Then from Corollary 3(i) we obtain that √ θ0 ≥ θ6 ≥ θ5 ≥ arccos z √ Let us consider two cases: (a) θ0 ≥ θ6 ≥ θ0 , (b) θ0 ≥ θ6 ≥ arccos z ˜ (a) We have h6 = H(y0 ; y1 , , y6 ) = H(y0 ; y1 , , y5 ) + f (θ6 ), H(y0 ; y1 , , y5 ) ≤ h5 = λ5 (ψ, θ0 ), ˜ ˜ f (θ6 ) ≤ f (θ0 ) ˜ Then h6 ≤ f (θ0 ) + λ5 (ψ, θ0 ) (b) In this case all θi ≤ θ0 ; i.e Y ⊂ C(e0 , θ0 ) Since H(y0 ; y1 , , y5 ) ≤ λ5 (ψ, θ0 ), √ it follows that h6 ≤ f (− z) + λ5 (ψ, θ0 ) √ ˜ f (θ6 ) ≤ f (− z), 27 THE KISSING NUMBER IN FOUR DIMENSIONS We have proved the following theorem Theorem Suppose n = 4, f ∈ Φ∗ (z), positive integer Then (i) h0 = f (1), √ z > t0 > z > 0, and N is a h1 = f (1) + f (−1); (ii) hm ≤ λm (ψ, θ0 ) ≤ λm (N, ψ, θ0 ) for ≤ m ≤ 5; √ √ ˜ (iii) h6 ≤ max {f (θ0 ) + λ5 (ψ, θ0 ), f (− z) + λ5 (ψ, θ0 )} ∀ θ0 ∈ [arccos z, θ0 ] 6-F Proof of Lemma B First we show that f4 ∈ Φ∗ (1/2) (see Fig 9) Indeed, the polynomial f4 has two roots on [−1, 1]: t1 = −t0 , t0 ≈ 0.60794, t2 = 1/2; f4 (t) ≤ for t ∈ [−t0 , 1/2], and f4 is a monotone decreasing function on the interval [−1, −t0 ] The last property holds because there are no zeros of the derivative f4 (t) on [−1, −t0 ] Thus, f4 ∈ Φ∗ (1/2) −1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 Figure The graph of the function f4 (t) We have t0 > 0.6058 Then Corollary 3(ii) gives μ ≤ For calculations of hm let us apply Theorem with ψ = arccos z = 60◦ , θ0 = arccos t0 ≈ 52.5588◦ We get h0 = f (1) = 18.774, h1 = f (1) + f (−1) = 24.48 H2 achieves its maximum at θ1 = 30◦ Then h2 = f (1) + 2f (− cos 30◦ ) ≈ 24.8644 For m = we have h3 = λ3 (60◦ , θ0 ) ≈ 24.8345 at θ3 = θ0 , θ1 = θ2 ≈ 30.0715◦ The polynomial H4 attains its maximum h4 ≈ 24.818 28 OLEG R MUSIN at the point with θ1 = θ2 ≈ 30.2310◦ , θ3 = θ4 ≈ 51.6765◦ , and h5 ≈ 24.6856 at α = 60◦ , θ1 ≈ 42.1569◦ , θ2 = θ4 = 32.3025◦ , θ3 = θ5 = θ0 √ ˜ ˜ Let θ0 = 50◦ We have f (50◦ ) ≈ 0.0906, arccos z = 45◦ , f (45◦ ) ≈ 0.4533, λ5 (60◦ , θ0 ) = h5 ≈ 24.6856, λ5 (60◦ , 50◦ ) ≈ 23.9181, ˜ ˜ h6 ≤ max { f (50◦ ) + h5 , f (45◦ ) + λ5 (60◦ , 50◦ ) } ≈ 24.7762 < h2 Thus hmax = h2 < 25 By (4.2), we have S(X) < 25M Concluding remarks This extension of the Delsarte method can be applied to other dimensions and spherical ψ-codes The most interesting application is a new proof for the Newton-Gregory problem, k(3) < 13 In dimension three computations of hm are technically much easier than for n = (see [26]) Let 2431 1287 18333 343 83 213 t f (t) = t − t + t + t − t − t + − 80 20 400 40 10 100 10 200 Then f ∈ Φ∗ (1/2), t0 ≈ 0.5907, μ(3, 1/2, f ) = 4, and hmax = h1 = 12.88 The (3) expansion of f in terms of Legendre polynomials Pk = Gk is f = P0 + 1.6P1 + 3.48P2 + 1.65P3 + 1.96P4 + 0.1P5 + 0.32P9 Since c0 = 1, ci ≥ 0, we have k(3) ≤ hmax = 12.88 < 13 Direct application of the method developed in this paper, presumably, could lead to some improvements in the upper bounds on kissing numbers in dimensions 9, 10, 16, 17, 18 given in [9, Table 1.5] (“Presumably” because the equality hmax = E is not proven yet.) In and 10 dimensions Table 1.5 gives: 306 ≤ k(9) ≤ 380, 500 ≤ k(10) ≤ 595 Our method gives: n=9: deg f = 11, E = h1 = 366.7822, t0 = 0.54; n = 10 : deg f = 11, E = h1 = 570.5240, t0 = 0.586 For these dimensions there is a good chance to prove that k(9) ≤ 366, k(10) ≤ 570 From the equality k(3) = 12, it follows that ϕ3 (13) < 60◦ The method gives ϕ3 (13) < 59.4◦ (deg f = 11) The lower bound on ϕ3 (13) is 57.1367◦ [16] Therefore, we have 57.1367◦ ≤ ϕ3 (13) < 59.4◦ 29 THE KISSING NUMBER IN FOUR DIMENSIONS By our approach it can be proven that ϕ4 (25) < 59.81◦ , ϕ4 (24) < 60.5◦ that can be proven by the same method as Theorem 4.) This improve the bounds: ϕ4 (25) < 60.79◦ , ϕ4 (24) < 61.65◦ [23] (cf [4]); ϕ4 (24) < 61.47◦ [4]; ϕ4 (25) < 60.5◦ , ϕ4 (24) < 61.41◦ [3] Now in these cases we have 57.4988◦ < ϕ4 (25) < 59.81◦ , 60◦ ≤ ϕ4 (24) < 60.5◦ However, for n = 5, 6, direct use of this extension of the Delsarte method does not give better upper bounds on k(n) than Odlyzko-Sloane’s bounds [27] It is an interesting challange to find better methods Appendix An algorithm for computation-suitable polynomials f (t) In this appendix we present an algorithm for computation “optimal”5 polynomials f such that f (t) is a monotone decreasing function on the interval [−1, −t0 ], and f (t) ≤ for t ∈ [−t0 , z], t0 > z ≥ This algorithm is based on our knowledge about optimal arrangement of points yi for given m Coefficients ck can be found via discretization and linear programming; such a method was employed by Odlyzko and Sloane [27] for the same purpose d We have a polynomial f represented in the form f (t) = + k=1 and the following constraints for f : (C1) ck ≥ 0, ≤ k ≤ d; (C2) f (a) > f (b) for −1 ≤ a < b ≤ −t0 ; (C3) (n) ck Gk (t) f (t) ≤ for −t0 ≤ t ≤ z We not know e0 where Hm attains its maximum; so for evaluation of hm let us use e0 = yc , where yc is the center of Δm All vertices yk of Δm are at the distance of ρm from yc , where cos ρm = (1 + (m − 1)z)/m When m = 2n − 2, Δm presumably is a regular (n − 1)-dimensional cross√ polytope.6 In this case cos ρm = z Let In = {1, , n} {2n − 2}, m ∈ In , bm = − cos ρm Then Hm (yc ) = f (1) + mf (bm ) If F0 is such that H(y0 ; Y ) ≤ E = F0 + f (1), then The long-standinding conjecture: The maximal kissing arrangment in four dimensions is unique up to isometry (in other words, is the “24-cell”), and ϕ4 (24) = 60◦ Open problem: Is it true that for given t0 , d this algorithm defines f with minimal hmax ? This is also an open problem 30 OLEG R MUSIN (C4) f (bm ) ≤ F0 /m, m ∈ In Note that E = F0 + + c1 + + cd = F0 + f (1) is a lower estimate of hmax A polynomial f that satisfies (C1-C4) and gives the minimal E can be found by the following: Algorithm Input: n, z, t0 , d, N Output: c1 , , cd , F0 , E First: replace (C2) and (C3) by a finite set of inequalities at the points aj = −1 + j, ≤ j ≤ N, = (1 + z)/N ; Second : Use linear programming to find F0 , c1 , , cd so as to minimize E−1 = d F0 + ck , subject to the constraints k=1 d ck ≥ 0, d (n) ≤ k ≤ d; (n) ck Gk (aj ) ≥ k=1 ck Gk (aj+1 ), k=1 d d (n) ck Gk (aj ) 1+ aj ∈ [−1, −t0 ]; ≤ 0, aj ∈ [−t0 , z]; (n) ck Gk (bm ) ≤ F0 /m, m ∈ In 1+ k=1 k=1 We note again that E ≤ hmax , and E = hmax only if hmax = Hm0 (yc ) for some m0 ∈ In University of Texas, Brownsville, Texas E-mail address: omusin@gmail.com References [1] K Anstreicher, The thirteen spheres: A new proof, Discrete and Computational Geometry 31 (2004), 613–625 [2] V V Arestov and A G Babenko, On Delsarte scheme of estimating the contact numbers, Proc of the Steklov Inst of Math 219 (1997), 36–65 [3] ——— , Estimates for the maximal value of the angular code distance for 24 and 25 points on the unit sphere in R4 , Math Notes 68 (2000), 419–435 [4] P G Boyvalenkov, D P Danev, and S P Bumova, Upper bounds on the minimum distance of spherical codes, IEEE Trans Inform Theory 42 (1996), 15761581 [5] ă ă K Boroczky, Packing of spheres in spaces of constant curvature, Acta Math Acad Sci Hung 32 (1978), 243–261 [6] ——— , The Newton-Gregory problem revisited, Proc Discrete Geometry, Marcel Dekker, New York, 2003, 103–110 [7] B C Carlson, Special Functions of Applied Mathematics, Academic Press, New York, 1977 [8] B Casselman, The difficulties of kissing in three dimensions, Notices Amer Math Soc 51 (2004), 884–885 THE KISSING NUMBER IN FOUR DIMENSIONS [9] 31 J H Conway and N J A Sloane, Sphere Packings, Lattices, and Groups (third edition), Springer-Verlag, New York, 1999 [10] H S M Coxeter, An upper bound for the number of equal nonoverlapping spheres that can touch another of the same size, Proc Sympos Pure Math 7, 53–71, A M S., Providence, RI (1963); Chap of Coxeter, Twelve Geometric Essays, Southern Illinois Press, Carbondale, IL, 1968 [11] L Danzer, Finite point-sets on S2 with minimum distance as large as possible, Discrete Math 60 (1986), 366 ă [12] L Danzer, B Grunbaum, and V Klee, Helly’s theorem and its relatives, Proc Sympos Pure Math 7, A M S., Providence, RI, (1963), 101–180 [13] Ph Delsarte, Bounds for unrestricted codes by linear programming, Philips Res Rep 27 (1972), 272–289 [14] Ph Delsarte, J M Goethals, and J J Seidel, Spherical codes and designs, Geom Dedicata (1977), 363–388 ´ [15] A Erdelyii (ed.), Higher Transcendental Function, McGraw-Hill Book Co., New York, Vol II, Chap XI (1953) ´ [16] L Fejes Toth, Lagerungen in der Ebene, auf der Kugel und in Raum, Springer-Verlag, New York, 1953; Russian translation, Moscow, 1958 [17] T Hales, The status of the Kepler conjecture, Mathematical Intelligencer 16 (1994), 47–58 [18] R Hoppe, Bemerkung der Redaction, Archiv Math Physik (Grunet) 56 (1874), 307–312 [19] W.-Y Hsiang, The geometry of spheres, in Differential Geometry (Shanghai, 1991), Word Sci Publ Co., River Edge, NJ, 1993, 92–107 [20] ——— , Least Action Principle of Crystal Formation of Dense Packing Type and Kepler’s Conjecture, World Sci., Publ Co., River Edge, NJ, 2001 [21] G A Kabatiansky and V I Levenshtein, Bounds for packings on a sphere and in space, Problems of Information Transmission 14 (1978), 1–17 [22] J Leech, The problem of the thirteen spheres, Math Gazette 41 (1956), 22–23 [23] V I Levenshtein, On bounds for packing in n-dimensional Euclidean space, Soviet Math Dokl 20 (1979), 417–421 [24] H Maehara, Isoperimetric theorem for spherical polygons and the problem of 13 spheres, Ryukyu Math J 14 (2001), 41–57 [25] O R Musin, The problem of the twenty-five spheres, Russian Math Surveys 58 (2003), 794–795 [26] ——— , The kissing problem in three dimensions, Discrete Comput Geom 35 (2006), 375–384 [27] A M Odlyzko and N J A Sloane, New bounds on the number of unit spheres that can touch a unit sphere in n dimensions, J Combinatorial Theory A26 (1979), 210–214 [28] F Pfender and G M Ziegler, Kissing numbers, sphere packings, and some unexpected proofs, Notices Amer Math Soc 51 (2004), 873–883 [29] I.J Schoenberg, Positive definite functions on spheres, Duke Math J (1942), 96108 ă [30] K Schutte and B L van der Waerden, Auf welcher Kugel haben 5,6,7,8 oder Punkte mit Mindestabstand Platz? Math Ann 123 (1951), 96124 ă [31] K Schutte and B L van der Waerden, Das Problem der dreizehn Kugeln, Math Ann 125 (1953), 325–334 32 OLEG R MUSIN [32] A D Wyner, Capabilities of bounded discrepancy decoding, Bell Systems Tech J 44 (1965), 1061–1122 (Received November 3, 2003) (Revised November 28, 2006) ... Mathematics, 168 (2008), 1–32 The kissing number in four dimensions By Oleg R Musin Abstract The kissing number problem asks for the maximal number k(n) of equal size nonoverlapping spheres in. .. method Introduction The kissing number k(n) is the highest number of equal nonoverlapping spheres in Rn that can touch another sphere of the same size In three dimensions the kissing number problem... and M Tagami: On optimal sets in Musin’s paper ? ?The kissing number in four dimensions? ?? in the Proceedings of the COE Workshop on Sphere Packings, November 1-5, 2004, in Fukuoka Japan) Now this claim